A plane is in a steep downward plunge. To not crash, captain Q pulls as much as he can. When the plane is at the lowest point of the plane's path the plane is in a circular arc of radius 450m. Captain Q experiences an acceleration of 5.0g ( so her weight feels like six times normal). How fast is the plane's going during this point?
To find the speed of the plane, we can use the concept of centripetal acceleration.
First, let's assume that the downward direction is positive. We can start by calculating the acceleration due to gravity experienced by Captain Q. The gravitational acceleration (g) is approximately 9.8 m/s^2.
Given that Captain Q experiences an acceleration of 5.0g, we can calculate the effective acceleration as follows:
Effective acceleration = 5.0g = 5.0 * 9.8 m/s^2 = 49 m/s^2
Since the plane is moving in a circular arc, the centripetal acceleration (ac) is given by the formula:
ac = v^2 / r
Where v is the velocity of the plane and r is the radius of the circle.
In this case, the radius is given as 450 meters. Rearranging the formula, we can solve for v:
v = √(ac * r)
Substituting the values:
v = √(49 m/s^2 * 450 m)
v = √(22050 m^2/s^2)
Therefore, the speed of the plane at the lowest point of its path is approximately 148.7 m/s.